1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

A power transmission line feeds input power at 2300 V to a srep down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :
A
50 A
B
45 A
C
35 A
D
25 A

Explanation

Given,

Primary voltage (VP) = 2300 V

Primary current (IP) = 5A

Secondary voltage (VS) = 230 V

efficiency ($$\eta $$) = 90%

We know,

Efficiency ($$\eta $$) = $${{{\mathop{\rm Sec}\nolimits} ondary\,Power} \over {\Pr imary\,\,Power}}$$

$$ \therefore $$   $$\eta $$ = $${{{P_S}} \over {{P_P}}}$$ = $${{{V_S}\,{I_S}} \over {{V_P}\,{I_P}}}$$

$$ \Rightarrow $$   0.9 = $${{230 \times {{\rm I}_S}} \over {2300 \times 5}}$$

$$ \Rightarrow $$   I = 45 A
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6 sin $$\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$$ then the maximum electric field associated with it is -
A
4.5 $$ \times $$ 104 N/C
B
4 $$ \times $$ 104 N/C
C
6 $$ \times $$ 104 N/C
D
3 $$ \times $$ 104 N/C

Explanation

E0 = B0 $$ \times $$ C

= 100 $$ \times $$ 10$$-$$6 $$ \times $$ 3 $$ \times $$ 108

= 3 $$ \times $$ 104 N/C

$$ \therefore $$  correct answer is 3 $$ \times $$ 104 N/C
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -
A
740 J
B
637.5 J
C
540 J
D
437.5 J

Explanation

$$L{{di} \over {dt}} = 25$$

$$L \times {{15} \over 1} = 25$$

$$L = {5 \over 3}H$$

$$\Delta E = {1 \over 2} \times {5 \over 3} \times ({25^2} - {10^2})$$

$$ = {5 \over 6} \times 525 = 437.5$$ J
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $$\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$$ The magnetic field $$\overrightarrow B $$(x,z, t) is given by $$-$$ (c is the velocity of light)
A
$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$
B
$${1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$
C
$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$
D
$${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$

Explanation

$$\overrightarrow E = 10\widehat j\cos \left[ {\left( {6\widehat i + 8\widehat k} \right).\left( {x\widehat i + z\widehat k} \right)} \right]$$

$$ = 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$$

$$ \therefore $$   $$\overrightarrow K = 6\widehat i + 8\widehat k;$$ direction of waves travel.

i.e., direction of 'c'.



$$ \therefore $$   Direction of $$\widehat B$$ will be along

$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$

Mag. of $$\overrightarrow B $$ will be along

$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$

Mag. of $$\overrightarrow B = {E \over C} = {{10} \over C}$$

$$ \therefore $$   $$\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}$$

Questions Asked from Alternating Current and Electromagnetic Induction

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